Page 571 - Mechatronics with Experiments

October 9, 2014 8:41 254mm×178mm
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JWST499-c07
JWST499-Cetinkunt
ELECTROHYDRAULIC MOTION CONTROL SYSTEMS 557
9
= 1.5 ⋅ 10 N∕m 2 bulk modulus (7.586)
3
V hose,pv = 1.0 ⋅ 10 −3 m volume between pump and valve (7.587)
3
V hose,VA = 1.0 ⋅ 10 −3 m volume between valve and A side (7.588)
3
V hose,VB = 1.0 ⋅ 10 −3 m volume between valve and B side (7.589)
2
A = 1.0 ⋅ 10 −3 m cylinder A side cross-section area (7.590)
A
2
A = 0.5 ⋅ 10 −3 m cylinder B side cross-section area (7.591)
B
l cyl = 1.0 m (7.592)
m = 10 kg piston and rod mass (7.593)
p
m = 990 kg load mass (7.594)
l
c = 100.0 cylinder damping (7.595)
p
2
F load (t) = (m + m ) ⋅ g = 1000 kg ⋅ 9.81 m∕s = 9810 N (7.596)
p
l
PID gains: (7.597)
K = 1.0 position sensor gain (7.598)
fb
K = 3.0 PID : proportional gain (7.599)
p
K = 0.0 PID : integral gain (7.600)
i
K = 0.0 PID : derivative gain (7.601)
d
Initial conditions: (7.602)
3
2
p (t ) = 101 ⋅ 10 N∕m pressures (7.603)
p 0
p (t ) = F load ∕A N∕m 2 (7.604)
A 0
a
3
p (t ) = 101 ⋅ 10 N ⋅ m 2 (7.605)
B 0
3
p (t ) = 101 ⋅ 10 N ⋅ m 2 (7.606)
T 0
i (t ) = 0.0A current (7.607)
v 0
x (t ) = 0.0m spool position (7.608)
v 0
̇ x (t ) = 0.0m∕s spool velocity (7.609)
v 0
y(t ) = 0.2m cylinder–piston position (7.610)
0
̇ y(t ) = 0.0m∕s cylinder–piston velocity (7.611)
0
The desired cylinder–piston is a trapeziodal position command,
y (t) = 0.2 ; t < 1.0 (7.612)
d
= 0.2 + 0.5 ⋅ (t − 1.0); 1.0 <= t < 2.0 (7.613)
= 0.7 ; 2.0 <= t < 3.0 (7.614)
= 0.7 − 0.5 ⋅ (t − 3.0); 3.0 <= t <= 4.0 (7.615)
= 0.2 ; 4.0 <= t (7.616)
The desired cylinder–piston velocity is
̇ y (t) = 0.0; t < 1.0 (7.617)
d
= 0.5; 1.0 <= t < 2.0 (7.618)
= 0.0; 2.0 <= t < 3.0 (7.619)
=−0.5; 3.0 <= t <= 4.0 (7.620)
= 0.0; 4.0 < t (7.621)